Resolved Problems of Linear Equations
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Introduction
A linear equation is a polynomial equation whose degree is 1,
because the highest degree of the monomials is one (the highest
power is 1, in x). Because of this, we'll have one solution,
if it has a solution, some don't.
In this section we are going to resolve exercises
with linear equations, the typical type of equations we'll
encounter in out day to day life. The solution requires a
lay out and a resolution of the equation with a unknown factor.
The exercises are in order of increasing difficulty,
starting with simple notions like
representing the double, triple and consecutive numbers
with algebra.
Although we suppose we know how to resolve this type of equations,
We remember that...
If we obtain an impossible equality,
there isn't a solution. For example, if
we obtain 1 = 0. This happens for example with
the equation x = x + 1, which is like saying a
number is equal to its consecutive,
which is false. Then, it's logical that the
equation doesn't have a solution.
If we obtain an equality that
always happens, either value is a solution,
and the solution is all the real numbers.
For example, if we obtain 0 = 0. This happens
with the equation x = x, which is the
same as saying a number is equal to itself,
which is always true.
If we still don't master linear equations
(with fractions, parenthesis, joint parenthesis...)
we can find resolved exercises in this section.
Resolved Problems step by step
Problem 1
Write the following expressions in algebraic form:
The double of a number x.
The triple of a number x.
The double of a number x plus 5.
The triple of a number x squared.
The three fourths parts of a number x.
Show solution
The double of a number is multiplying it by 2, therefore, the
double of x is 2x.
The triple of a number is multiplying it by 3, therefore
the triple of x is 3x.
The double of x is 2x, therefore if we add 5,
we have 2x + 5.
The triple of x is 3x, so when we square it we have
$$ (3x)^2 = 3^2 \cdot x^2 = 9x^2 $$
Note: we have used that the power of a
product is the product of powers.
The fourth part of x is
$$ \frac{1}{4} \cdot x = \frac{x}{4}$$
Therefore, three quarters of x are
$$ 3\cdot \frac{x}{4} = \frac{3x}{4} $$
Problem 2
Find the numbers, in each case, that apply:
Double it plus 5 is 35
-
Adding its consecutive number we obtain 51
-
Adding its double, its half and 15 we obtain 99
Its quarter part is 15
Show solution
The double of x is 2x, so we
have the equation
$$2x + 5 = 35$$
We resolve the equation:
$$2x = 35-5$$
$$2x = 30$$
$$x = \frac{30}{2}=15$$
Then, the number is 15.
If x is the number we're looking for, its consecutive
(the next one) is obtained by adding 1.
Then, adding x and its consecutive is
$$x + ( x + 1 ) = 51$$
We resolve the linear equation:
$$2x +1 = 51$$
$$2x = 50$$
$$x = \frac{50}{2} = 25$$
Then, the number is 25.
The double of x is 2x, and
half of x is x/2.
We have the equation:
$$x + 2x + \frac{x}{2} + 15 = 99$$
We resolve it:
$$3x + \frac{x}{2} = 99-15$$
$$3x + \frac{x}{2} = 84$$
We have to add fractions:
$$\frac{6}{2}x + \frac{1}{2}x = 84$$
$$\frac{7}{2}x = 84$$
$$x = \frac{84\cdot 2}{7}=24$$
Then, the number we are looking for is 24.
The fourth part of x is x/4. Then, the equation we
have to resolve is:
$$\frac{x}{4} = 15$$
$$x=15 \cdot 4 = 60$$
The number is 60.
Problem 3
Marta is 15 years old,
and that's a third part of her
mother’s age. How old is her mother?
Show solution
We call her mother's age x.
The third part of her mother's age is the same as Marta's, 15. Written mathematically:
$$\frac{x}{3} = 15$$
Therefore, her mother's age is x = 45.
Problem 4
How much does a rope measure if the three quarters of it measures 200 metres?
Show solution
If x is the length of the rope, we know
that three quarters is 200:
$$\frac{3x}{4} = 200$$
Therefore, the rope measures
$$x = \frac{4\cdot 200}{3} = 266.667\ meters $$
Problem 5
Obtain three consecutive numbers that added up gives 219.
Show solution
If x is the first number, the following
is x + 1, and the next is
$$(x + 1) + 1 = x + 2$$
We know that
$$ x + (x + 1) + (x + 2) = 219$$
$$3x + 3 = 219$$
$$3x = 219-3$$
$$3x = 216$$
$$x = \frac{216}{3} = 72$$
Then, the numbers are 72, 73 and 74.
Problem 6
We travel
a 1 km long road at 6km/h. How long will it take us to get to our destination?
Show solution
We know that the space travelled is the velocity multiplied by the
time, written mathematically,
$$ space\ travelled =velocity\cdot time $$
We call x = time:
We know that the space we need to travel
is 1 km, and the speed is 6 km/h.
We replace the values in the equation:
$$ 1 = 6\cdot x$$
$$ x = \frac{1}{6}$$
We will take x = 1/6 = 0.1667 hours in arriving, so,
$$ 0.1667\cdot 60 = 10.002 \ minutes$$
In reality, its exactly 10 minutes,
but we obtain 10.002 because we have approximated the value of 1/6.
Note: Don't forget to check the units of measurement.
If, for example, the space travelled was in metres,
we'd need to change it to kilometres
(or change the speed units).
Problem 7
James puts 25 dollars in his safe, which is a fourth
of the money in the safe to begin with. How much was in it to start with?
Show solution
We call x = the money that was in the safe
25 dollars are a fourth of what was in it. So,
$$ 25 = \frac{x}{4}$$
The solution of the equation is
$$ x = 25\cdot 4 = 100$$
In the safe there were 100$. Now there's.
$$ 100 + 25 = 125\ dollars$$
Problem 8
Anne's father is 5
years younger than her mother, and half her mother's age is 23. How old is Anne's father?
Show solution
x is the age of Anne's mother.
Half of the mother's age is 23, so
$$ \frac{x}{2} = 23$$
Anne's mother's age is
$$x = 23\cdot 2 = 46$$
Anne's father is 5 years younger than her mother, so, the father is 46 - 5 = 41 years old.
Problem 9
Stefanie is 16 years old and her two brothers are 2 and 3 years old.
How many years need to go by so that the double of the sum of her two brothers’
ages is the same as her age?
Show solution
We call x = years that need to go by
When these years pass, her brothers will be \(2+x\)
the youngest, and \(3 + x\) the oldest.
Also, Stefanie will be \(16 + x\) years old.
We want that
$$2( ( 2 + x )+( 3 + x ) ) = 16 + x$$
$$2( 2x + 5) = 16+x$$
$$4x +10 = 16 + x$$
$$4x-x = 16-10$$
$$3x = 6$$
$$x = \frac{6}{3}=2$$
So, two years need to go by.
Problem 10
Given a number, adding half of it, double it and triple it is 55.
What number is it?
Show solution
Let x the number we're looking for.
x/2 is its half
2x is its double
3x is its triple
The equation we have is
$$\frac{x}{2} + 2x + 3x = 55$$
$$\frac{x}{2} + 5x = 55$$
$$\frac{1}{2}x + \frac{10}{2}x = 55$$
$$\frac{11}{2}x = 55$$
$$x = \frac{55 \cdot 2}{11}=10$$
Therefore, the number is 10.
Problem 11
Vincent spends 20 dollars on a trousers and a shirt. He doesn't know
what each item costs, but he does know that the shirt is worth
two fifths of the trousers. How much do the trousers cost?
Show solution
If x is the price of the trousers, the shirt is
$$ \frac{2}{5x} $$
We know the total is 20 dollars, therefore,
$$x + \frac{2}{5}x = 20$$
We resolve the equation:
$$\frac{5}{5}x + \frac{2}{5}x = 20$$
$$\frac{7}{5}x = 20$$
$$x = \frac{20 \cdot 5}{7} = \frac{100}{7}=14,2857$$
The trousers are 14,29$.
Problem 12
The difference between two numbers is 17,
and the double of the smallest number is 26.
Which numbers are they?
And if 26 is double the biggest number, what numbers are they?
Show solution
First part:
Let x be the smallest number.
The difference is 17, so the biggest number is \(x + 17\).
The double of the smallest is 26, so
$$ 2x = 26 $$
Therefore, the numbers are
$$ x = \frac{26}{2} = 13 $$
and
$$ x + 17 = 13 + 17 = 30 $$
Second part:
Let x be the biggest number.
The difference is 17, so the smallest number is \(x - 17\).
The double of the biggest is 26, so
$$ 2x = 26 $$
The biggest number is 13 and the smallest is
$$ x - 17 = 13 - 17 = -4 $$
Problem 13
5 years ago, Mike was three times the age of his
cousin Joe, who is 15 years old. How many years need
to go by for Joe to have Mike's current age?
Show solution
x is Mike's current age
5 years ago, he was x - 5 years old,
and Joe
was 15 - 5 = 10.
5 years ago, Mike's age, x - 5,
was three times Joe's age:
$$ x - 5 = 3\cdot 10 $$
So, x = 30 + 5 = 35.
Mike is now 35 and Joe will be 35
years in 35 - 15 = 20 years time.
Problem 14
We have 3 fish tanks and 56 fishes.
The sizes of the fish tanks are small,
medium and large. The small one is half
the medium one and the big one is double
the medium.
We don't have a preference in how the fish are distributed but
we decide that in each of them there will be a proportional
amount of fish for the size.
How many fish will we put in each tank?
Show solution
Let's call:
x = number of fish in the medium size fishbowl
x/2 = the small fish tank and
2x = number of fish in the big tank
The total is 56 fish, so we have the following linear equation:
$$x + \frac{x}{2} + 2x = 56$$
We resolve it:
$$3x + \frac{x}{2} = 56$$
We have to add fractions:
$$\frac{6x}{2} + \frac{x}{2} = 56$$
$$\frac{7x}{2} = 56$$
The solution is
$$x = \frac{56\cdot 2}{7}= 16$$
Fish in the small tank: 16/2 = 8
Fish in the medium tank: 16
Fish in the big tank: 2·16 = 32
Problem 15
We want to distribute 510 sweets amongst 3 kids,
in a way that two of them have half of the sweets,
but one of these two has half of the others.
How many sweets will each child have?
Show solution
The quantity of sweets of two of them is
half of the total, so the other kid will have the other half:
$$ \frac{510}{2} = 255 $$
Now the remaining 255 need to be distributed
between the other two kids.
If x is the number of sweets
of one of the first kids (the one who has the most), then, the other kid
has half, so x/2.
Adding both kids' share gives 255, so we
have the equation
$$ x + \frac{x}{2} = 255$$
We add the fractions:
$$ \frac{3x}{2} = 255 $$
$$ x = \frac{255\cdot 2}{3} = 170$$
One has 170 sweets and the other 85.
The amount each kid will have is 255, 170 and 85.
Problem 16
One third part of the spoons of the house was in the dishwasher
and the rest in the drawer. But half the spoons in the drawer, 15,
are going to the table. How many spoons are there in the dishwasher?
Show solution
x = spoons in the house
In the dishwasher there was a third of the total:
$$\frac{x}{3}$$
The rest was in the drawer to start with. In the drawer there was:
$$ \frac{2x}{3}$$
Half of the spoons in the drawer were 15,
so there were 30 to start with.
So we obtain, by equalization, that
$$ \frac{2x}{3} = 30$$
In the house there are 45 spoons, 15 of which are in the dishwasher, 15 on the table and 15 in the drawer.
Problem 17
In a shop they sell a third of their products in
two days. The next day they receive from a warehouse
half of the quantity they sold, which is 15 units. How
many units did they sell in the first two
days? How many units are there in the shop after
they sell?
Show solution
Let be x the initial number of items.
In the first two days they sell
$$\frac{x}{3}$$
The next day they receive half of the sold quantity, 15. We have the equation:
$$\frac{1}{2}\cdot \frac{x}{3} = 15 $$
We have multiplied by 1/2 to obtain half.
We resolve the equation:
$$ \frac{x}{6} = 15$$
$$ x = 15 \cdot 6 = 90$$
In the shop there was 90 products.
In the first days they sell
$$\frac{x}{3} = \frac{90}{3} = 30$$
The quantity after they sell and restock is
$$ 90 - 30 + 15 = 75 \ products $$
Problem 18
Mark has 400 dollars and Rose has 350.
Both of them buy the same book.
After the purchase, Rose has five sixths of the money that Mark has left. Work out the price of the
book.
Show solution
x = price of the book
Money Mark has left : 400 - x
Money Rose has left : 350 - x
The money that Rose has left is 5/6 of what Mark has left, therefore,
$$ 350 - x = \frac{5}{6}( 400 - x )$$
We multiply by 6
$$ 2100 - 6x = 5(400-x) $$
$$ 2100 - 6x = 2000 -5x $$
$$ 2100 - 2000 = 6x -5x$$
$$ 100 = x$$
The price of the book is 100$.
Problem 19
Esther has three times the money Penny has and half of what John has. John gives Penny
and Esther 25 dollars each. Now, Esther has the same amount as John. How much did they
have to start with? And afterwards?
Show solution
x = Penny's initial money
Esther's initial money
was triple Penny's to start with, so Esther had 3x.
John's initial money was double Esther's, so John had
$$ 2\cdot 3x = 6x $$
He gives them money:
John has 50$ less, so he has
$$ 6x-50 $$
Penny has 25$ more, so Penny has
$$ x + 25 $$
Esther has 25$ more, so Esther has
$$ 3x + 25 $$
Now Esther has the same amount as John:
$$ 3x + 25 = 6x - 50 $$
The equation's solution is
$$ x = 25 $$
Initially they had:
Esther: 3·25 = 75$
Penny: 25$
John: 6·25 = 150€
Now they have:
Esther: 75 + 25 = 100$
Penny: 25 + 25 = 50$
John: 150 - 25 = 100$
Problem 20
In a house, the water deposit is at 2/7 its capacity.
Three people have showers: The first one uses one fifth of the quantity in the tank;
the second uses a third of what's in the tank and the third,
three fourths of the firsts' quantity.
Which is the capacity of the tank and how much water did the first two use is we know
that the third used 10 litres to shower?
Show solution
We call the capacity of the deposit x.
Before the showers, the deposit was at 2/7, so there are
$$\frac{2}{7}x \ liters$$
The first one uses one fifth of the quantity in the tank:
$$\frac{1}{5} \cdot \frac{2}{7}x = \frac{2}{35}x$$
The first one uses:
$$\frac{1}{5} \cdot \frac{2}{7}x$$
after that shower, there's
$$\frac{4}{5} \cdot \frac{2}{7}x = \frac{8}{35}x$$
left.
So, the second one uses
$$\frac{1}{3} \cdot \frac{8}{35}x = \frac{8}{105}x$$
The third uses three quarters of what the first one uses, so he consumes
$$\frac{3}{4} \cdot \frac{2}{35}x = \frac{3}{70}x$$
The third consumes 10L, so
$$\frac{3}{70}x = 10$$
Therefore,
$$x = \frac{700}{3}$$
The capacity of the deposit is 700/3 = 233.33 liters.
The first uses 2/35·700/3 = 40/3 = 13.33 liters.
The second uses 8/105·700/3 = 160/9 = 18.77 liters.
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